[LeetCode]121 Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]Output: 5max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]Output: 0In this case, no transaction is done, i.e. max profit = 0.

Kadane's Algorithm - Since no one has mentioned about this so far :) (In case if interviewer twists the input)

The logic to solve this problem is same as "max subarray problem" using Kadane's Algorithm. Since no body has mentioned this so far, I thought it's a good thing for everybody to know.

All the straight forward solution should work, but if the interviewer twists the question slightly by giving the difference array of prices, Ex: for {1, 7, 4, 11}, if he gives {0, 6, -3, 7}, you might end up being confused.

Here, the logic is to calculate the difference (maxCur += prices[i] - prices[i-1]) of the original array, and find a contiguous subarray giving maximum profit. If the difference falls below 0, reset it to zero.

    public int maxProfit(int[] prices) {        int maxCur = 0, maxSoFar = 0;        for(int i = 1; i < prices.length; i++) {            maxCur = Math.max(0, maxCur += prices[i] - prices[i-1]);            maxSoFar = Math.max(maxCur, maxSoFar);        }        return maxSoFar;    }

*maxCur = current maximum value

*maxSoFar = maximum value found so far

My Solution:

public class Solution {
    public int maxProfit(int[] prices) {
        int max = 0, sum = 0;
        for(int i = 1; i < prices.length; i++) {
            if(sum < 0)
                sum = prices[i] - prices[i - 1];
            else
                sum += prices[i] - prices[i - 1];
            if(max < sum)
                max = sum;
        }
        return max;
    }
}